My counterargument:
Intermolecular forces between molecules are either intact or broken. There is no in-between. Therefore, the change from intact to broken is instantaneous.
Applied heat energy increases the movement of particles in a substance, which in turn increases its temperature. When movement reaches a critical point, intermolecular force is overpowered, and a change of state has occurred.
The explanation that “energy goes to break the bonds, and not to increase temperature” makes no sense to me. In order to break the bond, the movement of the molecules needs to be intensified, which would increase the temperature of the water. That’s why I think that there should be no period of constant temperature.
My question: Where does that energy go? It’s not increasing temperature; so what is it doing?
Best Answer
Why does the temperature remain constant during a change in the state of matter? It doesn't exactly, as we see below, but it almost does under typical conditions where something called heterogeneous nucleation (that is, easy formation of molecular-scale clusters of the new phase) dominates. This is the case for water boiling right around 100°C at sea level in a standard kettle, for instance.
Two effects arise simultaneously. First, thermodynamically, one phase becomes the more stable phase at equilibrium. Second, kinetically, the creation of the new phase grows into an enormously fast process that moves quickly toward that equilibrium. Consequently, essentially any overheating is immediately absorbed to break bonds; conversely, any undercooling is immediately offset by energy released through bond formation. The process is fast—like exponential-function-applied-to-a-squared-term fast.
Both factors (thermodynamic and kinetic) are essential. Without the first, the driving pressure is absent to form regions of the new phase. Without the second, the formation process is slow or unpredictable (as in the case of room-temperature diamond decomposing to form more stable graphite, or a cup of microwaved water in a smooth, clean container that suddenly erupts partially into steam, or a bottle of ice from the freezer that quickly freezes when tapped).
Everything above repeats the standard explanation; the point of this answer is to present quantitative models, which are less often surveyed.
To address a point in the body of your question:
I don't think this is a great argument, as bonds are forming and breaking all the time in real materials at finite temperatures. Liquids have some fraction of broken bonds compared to the solid state, and the bonds themselves are fluctuating constantly. This certainly seems like an "in-between" to me.
In any case, your conceptual view can be shored up by replacing forces and bonds with the overall Gibbs free energy $G = H - TS$, where $H$ is the enthalpy (a measure of bonding, which Nature prefers), $T$ is the temperature, and $S$ is the entropy (a measure of broad possibilities, which Nature also prefers). Note that the entropy-containing term also includes the temperature.
The phase with the lowest Gibbs free energy is always the equilibrium state, as mediated by the tradeoff between increased bonding and increased entropy. In addition, the higher-entropy phase is always the higher-temperature stable phase. For the typical single-component system discussed here, only at one temperature can two phases have the same $G$: the phase-change temperature. This is part of why we seem to see a sudden transition at this temperature.
We can visualize this behavior in the following image. The lowest curve is the most stable. On the left side, the curves are more shallow because entropy $S$ (the negative slope) decreases toward zero with decreasing $T$. But the larger-entropy phases always win out with increasing temperature because of the $-TS$ term. (If the liquid curve were a little higher, we'd have a material such as carbon dioxide that doesn't exhibit an equilibrium liquid phase—at least not at atmospheric pressure—and sublimates directly from the solid to the gas state upon heating.)
Now let's get into why the temperature seems to stay constant during a phase change. It doesn't, in fact, but it is true that a only a very slight amount of undercooling or overheating can be sufficient to trigger a complete rapid transformation from the less stable phase to the more stable phase. This discussion follows Porter & Easterling, Phase Transformations in Metals and Alloys, but does not assume any particular material class.
Consider slight overheating past one of the transition points in the diagram above. The volumetric driving force for the phase transformation is the reduction in Gibbs free energy, $\Delta G=\Delta H - T\Delta S$, where $\Delta$ compares the state values for the two phases (i.e., $\Delta H=H_1-H_2$, for example, where phase 1 is more stable than phase 2 in this case). Close to the intersection, $\Delta H$ is just the latent heat $L$ of the transformation. In addition, we know that $\Delta G =0$ at the transformation temperature $T_t$, so $$\Delta G = L\left(1-\frac{T}{T_t}\right)=\frac{L\Delta T}{T_t}.$$ Thus, the driving force $|\Delta G|$ for transformation is modeled as increasing linearly with increasing overheating or undercooling $\Delta T$.
Now consider the benefits and penalties of nucleating a new phase in the form of a spherical molecular cluster with radius $r$. The benefit is the reduction in Gibbs free energy $\frac{4}{3}\pi r^3\Delta G$. The penalty is the new interface area $4\pi r^2\gamma$, where $\gamma$ is the surface tension of the new phase with respect to the old phase.
Clusters form and dissipate constantly from random particle motion; for just a few molecules, $r^3\ll r^2$, and if the overheating/underheating is low, there's no energetic driving force for further growth, and the cluster is likely to just come apart. But a larger temperature excursion alters this balance substantially.
A cluster becomes stable (at radius $r^\star$) when an increase in size starts to produce a net decrease in energy, i.e., when a maximum exists and the derivative is zero: $$\frac{d}{dr}\left(4\pi r^2\gamma-\frac{4}{3}\pi r^3\Delta G\right)=0.$$ Solving this equation, we find that at the stable radius, the activation energy or energy "hump" that needs to be overcome is $$\Delta G^\star=\frac{16\pi\gamma^3}{3(\Delta G)^2}.$$ The corresponding expected volumetric nucleation rate $N$ from Arrhenius theory is then $$N=f\exp\left(-\frac{\Delta G^\star}{kT}\right)=f\exp\left(-\frac{16\pi\gamma^3}{3(\Delta G)^2kT}\right)=f\exp\left(-\frac{A\gamma^3}{(\Delta T)^2}\right),$$ where $k$ is Boltzmann's constant and where $f$ is a characteristic (very high) vibration frequency in the material; Porter & Easterling use $f\approx 10^{11}\,\mathrm{Hz}$, for example. To simplify the discussion, take $A$ to be a constant that incorporates everything except two parameters of interest: the surface tension difference $\gamma$ and the overheating/undercooling $\Delta T$.
The surface tension $\gamma$ would be important if every cluster interface routinely separated only the old phase from the new phase. However, $\gamma$ can be much lower if the cluster initiates on a particle of dirt or a crack in a container wall that has a high energy anyway because of unsatisfied bonds. The former case is called homogenous nucleation, and the latter—which we often see in our dirty, defect-filled world—is called heterogenous nucleation.
For small $A\gamma^3$, try plotting the nucleation rate $N\sim \exp\left(-\frac{A\gamma^3}{(\Delta T)^2}\right)$. You'll see a tremendously rapid increase in nucleation for slight overheating/undercooling because we're applying the exponential function to a squared temperature difference:
The kinetics of the process are so strong that essentially any overheating is immediately shunted into breaking bonds to form the now-favored higher-temperature phase. (Conversely, any undercooling past the phase transition temperature is immediately inundated with energy released from bond formation to provide the now-favored lower-temperature phase.)
This is ultimately why phase transitions seem to occur entirely at a single well-defined temperature. That's not quite the case, as shown above, but it can be a reasonably good approximation.